NL
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
22 tháng 2 2019
\(\dfrac{x-90}{10}+\dfrac{x-76}{12}=\dfrac{x-58}{14}+\dfrac{x-36}{16}+\dfrac{x-15}{17}=15\)
\(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)=\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)\)\(\Leftrightarrow\dfrac{x-100}{10}+\dfrac{x-100}{12}=\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}-\dfrac{1}{14}-\dfrac{1}{16}-\dfrac{1}{17}\right)=0\)
\(\Leftrightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
17 tháng 6 2022
b: \(\Leftrightarrow4x^2-8x+4=x^2+2x+1+3\left(x^2+x-6\right)\)
\(\Leftrightarrow3x^2-10x+3=3x^2+3x-18\)
=>-13x=-21
hay x=21/13
c: \(\Leftrightarrow\left(\dfrac{x-90}{10}-1\right)+\left(\dfrac{x-76}{12}-2\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-15}{17}-5\right)=0\)
=>x-100=0
hay x=100
21 tháng 1 2018
a) \(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\)
\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}=\dfrac{x}{6}=\dfrac{6x}{6}\)
\(\Leftrightarrow2x-3\left(2x+1\right)=x-6x\)
\(\Leftrightarrow2x-6x-3=x-6x\)
\(\Leftrightarrow2x-6x-x+6x=3\)
\(\Leftrightarrow x=3\)
\(S=\left\{3\right\}\)
b) \(\dfrac{2+x}{5}-0,5x=\dfrac{1-2x}{4}+0,25\)
\(\Leftrightarrow\dfrac{4\left(2+x\right)}{20}-\dfrac{10x}{20}=\dfrac{5\left(1-2x\right)}{20}+\dfrac{5}{20}\)
\(\Leftrightarrow4\left(2+x\right)-10x=5\left(1-2x\right)+5\)
\(\Leftrightarrow8+4x-10x=5-10x+5\)
\(\Leftrightarrow4x-10x+10x=5+5-8\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
\(S=\left\{\dfrac{1}{2}\right\}\)
6 tháng 1 2018
\(\dfrac{x+1}{94}+\dfrac{x+2}{93}+\dfrac{x+3}{92}=\dfrac{x+4}{91}+\dfrac{x+5}{90}+\dfrac{x+6}{89}\)
\(\Rightarrow\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)
\(\Rightarrow\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}=\dfrac{x+95}{91}+\dfrac{x+95}{90}+\dfrac{x+95}{89}\)
\(\Rightarrow\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)
\(\Rightarrow\left(x+95\right)\left(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\right)=0\)
Vì \(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\ne0\) nên \(x+95=0\Leftrightarrow x=-95\)
6 tháng 1 2018
Mk làm luôn nhé , không chép lại đề đâu !!! Ahihi
\(\dfrac{x+1}{94}+1+\dfrac{x+2}{93}+1+\dfrac{x+3}{92}+1=\dfrac{x+4}{91}+1+\dfrac{x+5}{90}+1+\dfrac{x+6}{89}+1\)⇔\(\dfrac{x+95}{94}+\dfrac{x+95}{93}+\dfrac{x+95}{92}-\dfrac{x+95}{91}-\dfrac{x+95}{90}-\dfrac{x+95}{89}=0\)
⇔ \(\left(x+95\right)\)\(\left(\dfrac{1}{94}+\dfrac{1}{93}+\dfrac{1}{92}-\dfrac{1}{91}-\dfrac{1}{90}-\dfrac{1}{89}\right)\) = 0
⇔\(x+95=0\)
⇔ \(x=-95\)
Vậy , ......
31 tháng 12 2022
a: \(\Leftrightarrow4x+4+9\left(2x+1\right)=2\left(5x+3\right)+12x+7\)
=>4x+4+18x+9=10x+6+12x+7
=>22x+13=22x+13(luôn đúng)
b: \(\Leftrightarrow\left(\dfrac{x+1}{94}+1\right)+\left(\dfrac{x+2}{93}+1\right)+\left(\dfrac{x+3}{92}+1\right)=\left(\dfrac{x+4}{91}+1\right)+\left(\dfrac{x+5}{90}+1\right)+\left(\dfrac{x+6}{89}+1\right)\)
=>x+95=0
=>x=-95
TA
30 tháng 1 2018
1a)\(\dfrac{x-90}{10}-1+\dfrac{x-76}{12}-2+\dfrac{x-58}{14}-3+\dfrac{x-36}{16}-4+\dfrac{x-15}{17}-5=0\)
=> \(\dfrac{x-100}{10}+\dfrac{x-100}{12}+\dfrac{x-100}{14}+\dfrac{x-100}{16}+\dfrac{x-100}{17}=0\)
=>\(\left(x-100\right)\left(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\right)=0\)
=> x=100( vi \(\dfrac{1}{10}+\dfrac{1}{12}+\dfrac{1}{14}+\dfrac{1}{16}+\dfrac{1}{17}\ne0\)
b) \(\dfrac{x-5}{2012}-1+\dfrac{x-4}{2013}-1=\dfrac{x-3}{2014}-1+\dfrac{x-2}{2015}-1\)
=> \(\dfrac{x-2017}{2012}+\dfrac{x-2017}{2013}-\dfrac{x-2017}{2014}-\dfrac{x-2017}{2015}=0\)
=>(x-2017).\(\left(\dfrac{1}{2012}+\dfrac{1}{2013}-\dfrac{1}{2014}-\dfrac{1}{2015}\right)=0\)
=> x=2015(vi .....................................................≠0)
2)
SG
5
KT
22 tháng 4 2017
\(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\\ \Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\\ \Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}=\dfrac{x+10}{7}+\dfrac{x+10}{6}\\ \Leftrightarrow\left(x+10\right).\dfrac{1}{9}+\left(x+10\right).\dfrac{1}{8}-\left(x+10\right).\dfrac{1}{7}-\left(x+10\right).\dfrac{1}{6}=0\\ \Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)
vì \(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\ne0\)
nên \(x+10=0\Rightarrow x=-10\)
vậy phương trình có tập nghiệm là S={-10}
SG
2
24 tháng 4 2017
\(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
\(\Leftrightarrow\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+4}{96}+1\right)=\left(\dfrac{x+6}{94}+1\right)+\left(\dfrac{x+8}{92}+1\right)\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}+\dfrac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)
Vì : \(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\ne0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
Vậy ...............
24 tháng 4 2017
Võ Đông Anh Tuấn Thế nhỡ \(\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\) thì x=-100 có phải nghiệm không
\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) ( x # 0 ; x # 6)
⇔ \(\dfrac{90\left(x-6\right)-36x}{x\left(x-6\right)}=\dfrac{2x\left(x-6\right)}{x\left(x-6\right)}\)
⇔ 90x - 540 - 36x = 2x2 - 12x
⇔-2x2 + 66x - 540 = 0
⇔ -2( x2 - 33x +270 ) = 0
⇔ x2 - 18x - 15x + 270 = 0
⇔ x( x - 18) - 15( x - 18) = 0
⇔ ( x - 18)( x - 15) = 0
⇔ x = 18 ( TM) hoac x = 15 ( TM)
KL........